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bazaltina [42]
3 years ago
9

PLZ HELP FAST! Which two statements help explain why digital storage of data is so reliable?

Physics
2 answers:
Shkiper50 [21]3 years ago
3 0

Answer:

C and D

Explanation:

But really, You should be able to answer this with the tech knowledge of a tomato.  You're given four answers, and are to choose which two are explain the reliability of digital storage.

The first two describe bad nasty things, the second two describe beneficial things.

So logically....

Aleonysh [2.5K]3 years ago
3 0

Answer:

C and d

Explanation:

A P E X

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If light moves at a speed of 299,792,458 m/s, how long will it take light to move a distance of 1,000 km
astra-53 [7]

Answer:

change 1000km into Metre

1000km=1000000 M

speed= distance/time

time=distance/speed

time=1000000/299792458

time= 0.0033second

6 0
3 years ago
A bicycle has a momentum of 23.4
Snowcat [4.5K]
  • Momentum=23.4kgm/s
  • Velocity=2m/s

\\ \rm\longmapsto Momentum=Mass\times velocity

\\ \rm\longmapsto Mass=\dfrac{Momentum}{Velocity}

\\ \rm\longmapsto Mass=\dfrac{23.4}{2}

\\ \rm\longmapsto Mass=11.7kg

4 0
3 years ago
Assume that the polymer material has a constant refractive index of 1.5. For light of 600nm wavelength at normal incidence, what
yaroslaw [1]

Answer:

Minimum thickness will be 100 nm

Explanation:

We have given refractive index is n = 1.5

Wavelength of the light incidence \lambda= 600 nm

We have to find the smallest thickness of the film so that there will be minimum light reflect

For minimum thickness of non reflecting film

t=\frac{\lambda }{4n} , here t is thickness, \lambda is wavelength and n is refractive index

Putting all values t=\frac{600}{4\times 1.5}=100nm

So minimum thickness will be 100 nm

8 0
3 years ago
What must be true about two objects if heat is flowing between them?
netineya [11]

Answer:

The objects must be different temperatures.

Explanation:

For heat to flow between two objects, heat must be flowing between them. The thermal gradient allows for the flow of heat. Heat is a form of energy that is dissipated from one place to another based on temperature difference.

Temperature is the degree of hotness or coldness of body. It is power by heat energy between two bodies.

Heat generally flows from a body at high temperature to one at low temperature. When thermal equilibrium is established and both bodies have the same temperature, heat will stop to flow.

8 0
3 years ago
Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
shusha [124]

Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

                              F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2}  )

- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

8 0
4 years ago
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